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I understand how to connect a monstable multivibrator to be in the non-retriggerable mode. Just loopback the /Q to the falling edge input or the Q to the rising edge output) as in the following picture (taken from the MC14528B datasheet).

enter image description here

Now I think I understand this can be used as a divide by two counter circuit. If I feed it 1kHz square wave on the A pin, i should get a 500Hz square wave on the Q output pin. This would give the following timing diagram:

enter image description here

I do not understand why the feedback of the /Q output into the falling edge trigger input (B) enables this behavior? Can anybody provide more insight on why this loop back suppresses the action of the rising edge on the A port at 1ms?

Francis Poirier
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    In short: it doesn't. What makes you think it does? It will if the monostable delay time is longer but you haven't furnished your question with all the details. – Andy aka Sep 19 '20 at 14:09
  • Why i thought this would work is because someone where i work made a quadrature pulse generator using two monostable multivibrators (one set to trigger on the positive edge and the other on the negative edge) and an input frequency of 2f. I don't know how he did it but after reading some datasheets i thought the non-retriggerable mode was it. In that particular situation, having a RC time constant slightly longer than the input signal period wouldn't work because in the long run, the signals would drift out of sync no? Is there an element missing to make this divide by 2 counter? – Francis Poirier Sep 19 '20 at 15:58
  • I just realized signals would stay in sync with long RC time constant, it's just that the duty cycle would be slightly more than 50%. Thank you very much for your help! I think maybe this question should be closed as it is confusing and unlikely to be useful to someone else? – Francis Poirier Sep 19 '20 at 16:17
  • upvote for your attitude – jsotola Sep 19 '20 at 17:39

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